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A circle with center O and radius 5 is shown in the xy-plane. Lines that intersect the circle in 2 points include which of the following ?
Let's start by understanding what we're looking for. We have a circle with center O and radius 5. We need to figure out which of the three given lines will intersect this circle at exactly 2 points.
Think about it this way: imagine throwing a stick (the line) near a hula hoop (the circle). If the stick passes through the hoop, it will touch the hoop at exactly 2 points. But if the stick is too far away, it won't touch the hoop at all. And if the stick passes exactly through the center, it still touches at 2 points, but if it just barely grazes the edge, it touches at only 1 point.
So we're looking for lines that pass through the circle (not just graze it, and not miss it entirely).
Process Skill: TRANSLATE - Converting the geometric requirement into a distance comparison problem
Here's the key insight: a line intersects a circle in exactly 2 points when the shortest distance from the center of the circle to the line is less than the radius.
Why does this make sense? Picture the center of the circle and draw the shortest possible line from that center to your line - this will always be a perpendicular line. If this perpendicular distance is:
- Less than the radius: the line cuts through the circle (2 intersection points)
- Equal to the radius: the line just touches the circle (1 intersection point - tangent)
- Greater than the radius: the line misses the circle entirely (0 intersection points)
Since our circle has radius 5, we need the perpendicular distance from center O to each line to be less than 5.
Since the problem doesn't specify where center O is located, we can assume it's at the origin \(\mathrm{(0,0)}\) - this is standard for GMAT problems unless stated otherwise.
For a line in the form \(\mathrm{ax + by + c = 0}\), the distance from point \(\mathrm{(x_0, y_0)}\) to the line is:
\(\mathrm{Distance = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}}\)
Let's convert each line to standard form and calculate:
Line I: \(\mathrm{y = -x + 1}\)
Rearranging: \(\mathrm{x + y - 1 = 0}\)
So \(\mathrm{a = 1, b = 1, c = -1}\)
Distance from \(\mathrm{(0,0) = \frac{|1(0) + 1(0) + (-1)|}{\sqrt{1^2 + 1^2}} = \frac{|−1|}{\sqrt{2}} = \frac{1}{\sqrt{2}} ≈ 0.71}\)
Line II: \(\mathrm{y = 2x + 1}\)
Rearranging: \(\mathrm{2x - y + 1 = 0}\)
So \(\mathrm{a = 2, b = -1, c = 1}\)
Distance from \(\mathrm{(0,0) = \frac{|2(0) + (-1)(0) + 1|}{\sqrt{2^2 + (-1)^2}} = \frac{|1|}{\sqrt{5}} = \frac{1}{\sqrt{5}} ≈ 0.45}\)
Line III: \(\mathrm{y = \frac{1}{2}x - 6}\)
Rearranging: \(\mathrm{\frac{1}{2}x - y - 6 = 0}\), or multiplying by 2: \(\mathrm{x - 2y - 12 = 0}\)
So \(\mathrm{a = 1, b = -2, c = -12}\)
Distance from \(\mathrm{(0,0) = \frac{|1(0) + (-2)(0) + (-12)|}{\sqrt{1^2 + (-2)^2}} = \frac{|−12|}{\sqrt{5}} = \frac{12}{\sqrt{5}} ≈ 5.37}\)
Now we compare each distance to our radius of 5:
- Line I: Distance ≈ 0.71 < 5 ✓ (intersects circle in 2 points)
- Line II: Distance ≈ 0.45 < 5 ✓ (intersects circle in 2 points)
- Line III: Distance ≈ 5.37 > 5 ✗ (does not intersect circle)
Therefore, only lines I and II intersect the circle in 2 points.
Process Skill: APPLY CONSTRAINTS - Systematically checking each option against our geometric criterion
Lines I and II both intersect the circle in 2 points, while line III does not intersect the circle at all.
The answer is C. I and II only.
1. Misunderstanding the circle's center location
Students often assume the circle is centered at some other point rather than the origin \(\mathrm{(0,0)}\). Since the problem states "center O" without specifying coordinates, students might place it at \(\mathrm{(5,0)}\) or another location, leading to incorrect distance calculations throughout the solution.
2. Confusing intersection conditions
Students frequently mix up when a line intersects a circle in 2 points versus 1 point versus 0 points. They might think that if the distance equals the radius, the line intersects in 2 points, when actually this creates a tangent (1 intersection point). The correct condition is that distance < radius for 2 intersection points.
3. Forgetting to convert lines to standard form
Many students attempt to apply the point-to-line distance formula directly to the slope-intercept form \(\mathrm{(y = mx + b)}\) without first converting to standard form \(\mathrm{(ax + by + c = 0)}\). This leads to incorrect identification of coefficients a, b, and c in the distance formula.
1. Arithmetic errors in distance calculations
Students commonly make calculation mistakes when computing distances, especially with square roots. For example, when calculating \(\frac{1}{\sqrt{5}}\), they might incorrectly approximate it or make errors in the fraction arithmetic, leading to wrong comparisons with the radius of 5.
2. Sign errors when converting to standard form
When rearranging equations like \(\mathrm{y = -x + 1}\) to standard form \(\mathrm{x + y - 1 = 0}\), students frequently make sign errors. They might write \(\mathrm{x + y + 1 = 0}\) instead, which changes the entire distance calculation and leads to incorrect conclusions about which lines intersect the circle.
3. Incorrect coefficient identification
Even when students correctly convert to standard form, they often misidentify the values of a, b, and c. For instance, with \(\mathrm{2x - y + 1 = 0}\), they might confuse which coefficient is negative or mix up the order when substituting into the distance formula.
1. Misreading the Roman numeral combinations
After correctly determining that lines I and II intersect the circle while line III does not, students sometimes select the wrong answer choice. They might choose "A. I only" if they focus on the first correct line, or get confused by the Roman numeral notation and select an incorrect combination.